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Author: Tinku Tara

WeinGarten-Equation-r-u-r-u-E-r-u-r-v-F-r-v-r-v-G-N-N-1-N-N-u-N-u-N-N-N-u-0-N-N-v-N-v-N-N-N-v-0-N-N-0-N-N-N

Question Number 225768 by Lara2440 last updated on 09/Nov/25 $$\mathrm{WeinGarten}\:\mathrm{Equation} \\ $$$$\:\boldsymbol{\mathrm{r}}_{{u}} \centerdot\boldsymbol{\mathrm{r}}_{{u}} ={E}\:,\:\boldsymbol{\mathrm{r}}_{{u}} \centerdot\boldsymbol{\mathrm{r}}_{{v}} ={F}\:,\:\boldsymbol{\mathrm{r}}_{{v}} \centerdot\boldsymbol{\mathrm{r}}_{{v}} ={G} \\ $$$$\hat {\boldsymbol{\mathrm{N}}}\centerdot\hat {\boldsymbol{\mathrm{N}}}=\mathrm{1} \\ $$$$\frac{\partial\left(\hat {\boldsymbol{\mathrm{N}}}\centerdot\hat…

Question-225740

Question Number 225740 by fantastic last updated on 08/Nov/25 Answered by fantastic last updated on 08/Nov/25 $$\gamma=\mathrm{sin}^{−\mathrm{1}} \left(\frac{{r}}{\mathrm{1}−{r}}\right) \\ $$$$\alpha=\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{3}{r}}{\mathrm{1}−{r}}\right) \\ $$$$\beta=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\mathrm{2}{r}}{\mathrm{1}}\right) \\…

Question-225703

Question Number 225703 by ajfour last updated on 07/Nov/25 Answered by mahdipoor last updated on 07/Nov/25 $$\mathrm{2}×\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} =\left(\mathrm{R}+\mathrm{r}\right)^{\mathrm{2}} \Rightarrow\mathrm{r}=\mathrm{R}\left(\mathrm{3}−\sqrt{\mathrm{8}}\right) \\ $$$$\mathrm{center}\:\mathrm{of}\:\mathrm{balls}\:\left(\mathrm{r}\right)\:\mathrm{in}\:\mathrm{R}_{\mathrm{c}} =\mathrm{R}−\mathrm{r}\:\: \\ $$$$\theta_{\mathrm{each}\:\mathrm{ball}} =\mathrm{2tan}^{−\mathrm{1}}…